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4A.13

Posted: Wed Feb 05, 2020 4:47 pm
by vanessas0123
A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 8C. In a subsequent experiment, 100.0 mL of 0.200MHBr(aq) and 100.0 mL of 0.200M KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 8C. What is the change in the internal energy?

For the second part where you plug in the Ccal calculated into the q=-Ccal(delta T) in order to find q or delta U, why is there a negative sign in front of Ccal?

Re: 4A.13

Posted: Wed Feb 05, 2020 7:32 pm
by Adriana_4F
The heat lost by the reaction is equal to the heat gained by the calorimeter (qcal=-qrxn). In order to calculate the change in internal energy of the reaction, you need to know the amount of heat that was lost by the reaction, so you do the opposite of that equation (qrxn=-qcal => qrxn=-Ccal(delta T))